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Q.
The equation $16x^2 - 3y^2 - 32x + 12y - 44 = 0$ represents a hyperbola
Conic Sections
Solution:
The given equation can be written as
$16\left(x^{2}-2x\right)-3\left(y^{2} -4y\right)-44=0$
i.e., $16\left(x-1\right)^{2}-3\left(y-2\right)^{2} = 44+16-12$
is given by $''T = S_{1}''$ i.e.,
$ \Rightarrow \frac{\left(x-1\right)^{2}}{3} - \frac{\left(y-2\right)^{2}}{16} = 1$
which is a hyperbola whose eccentricity is given by
$ b^{2}= a^{2}\left(e^{2}-1\right)$
$ \Rightarrow 16= 3\left(e^{2}-1\right)$
$ \Rightarrow e^{2} -1 = \frac{16}{13} $
$ \Rightarrow e^{2} = 1+\frac{16}{3} = \frac{19}{3} $
$ \Rightarrow e = \sqrt{19/3 }$